I want to show that $\dim V = \dim \phi(V)+\dim \ker \phi$.
I know this proof can be found in any linear algebra textbook. However, my question is not exactly about the proof, but on a statement I assumed to be true while attempeting to prove it myself.
In order to prove this, let $\{{e_1,\dots,e_p}\}$ be a basis of $\phi^{-1}\{0\}$ and $\{{e_1,\dots,e_p},{\bar{e_1},\dots,\bar{e_q}}\}$ be a basis of $V$. Then, $\dim \ker \phi=p$ and $\dim V = p+q$.
So it suffices to show that $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ is a basis of $\phi(V)$.
I managed to show that $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ spans $\phi(V)$. But I still need to prove that the set $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ is linearly independent. This is how I approached the problem:
Consider $a^{i}e_i + b^{i}\bar{e_i}=0$. Assume the sum is over the whole sets, i.e, the first goes up to $p$ and the second up to $q$. Since $\{{\bar{e_1}},e_i\}$ is a basis of $V$, $a^{i}=b^{i}=0$ for all $i$.
Since $\phi(0)=0$ and $\phi$ is linear: $$ \phi(a^{i}e_i + b^{i}\bar{e_i})=\phi(0)=0 \Rightarrow a^{i}\phi(e_i) + b^{i}\phi(\bar{e_i})=0 $$
We also know that $\{{e_1,\dots,e_p},{\bar{e_1},\dots,\bar{e_q}}\}\subset \phi^{-1}(0)$, so $\phi(e_i)=0$ for all $i$. Thus, we have: $$ b^{i}\phi(\bar{e_i})=0 $$ and all the $b^{i}$'s are $0$.
However, I know I should start with this last equation and then conclude that all the $b^{i}$'s are $0$. I can reverse my reasoning and show that this also works in the other way round.
To do so, I would have to assume that $\phi(a^{i}e_i + b^{i}\bar{e_i})=\phi(0)$ implies $a^{i}e_i + b^{i}\bar{e_i}=0$. But I don't see how the second statement follows from the first.
